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Kepler's Law in Schwarzchild metric

  1. Mar 25, 2016 #1

    Fek

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    1. The problem statement, all variables and given/known data

    Show Kepler's Third Law holds for circular Schwarzchild orbits.

    2. Relevant equations
    3. The attempt at a solution

    Setting [itex] r' = 0 , \theta' = 0 [/itex] and [itex] \theta = \pi / 2 [/itex] , where the derivatives are with respect to the variable [itex] \lambda [/itex] and setting c = 1 the Lagrangian is:

    [itex] L = (1 - r_s / r) \dot t^2 - r^2\dot\phi^2 [/itex]

    E-L equation for r:

    $$\frac{r_s} {r^2} \dot t^2 - 2r \dot\phi^2 = 0$$

    Therefore $$d\phi / dt = GM/r $$ , the solution.

    However I'm struggling to understand why t here corresponds to the proper time at infinity, but r corresponds to the radius of the ciruclar orbit?
     
  2. jcsd
  3. Mar 30, 2016 #2
    Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
     
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